2016 AMC 10B Problem 12

Below is the professionally curated solution for Problem 12 of the 2016 AMC 10B, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2016 AMC 10B solutions, or check the answer key.

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Concepts:basic probability complementary counting parity

Difficulty rating: 960

12.

Two different numbers are selected at random from $\{1, 2, 3, 4, 5\}$ and multiplied together. What is the probability that the product is even?

$\ 0.2$

$\ 0.4$

$\ 0.5$

$\ 0.7$

$\ 0.8$

Solution:

The product is odd if and only if both numbers are odd. There are $\binom 32=3$ ways to do this out of a possible $\binom 52=10$ ways. This makes the probability of it being odd equal to $\frac 3{10} = 0.3.$ This means the probability it is even is $1-0.3=0.7.$

Thus, the correct answer is D .

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